See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The iodoform test is positive for compounds that have the structural feature of a methyl ketone (CH3CO-) or compounds that can be oxidized to a methyl ketone, including ethanol (CH3CH2OH) and acetaldehyde (CH3CHO). The reagent is I2/NaOH. Step 1: Identify the structural requirement. A compound gives the iodoform test if it contains the CH3CH(OH)- group or CH3CO- group, because NaOH/I2 first oxidizes secondary alcohols with a methyl group adjacent to the OH, or directly acts on methyl ketones/acetaldehyde. Step 2: Analyze option (a) CH3CH2OH (ethanol). Ethanol has the structure CH3-CH2-OH. It contains the CH3CH(OH)- pattern (the CH3 group attached to a carbon bearing OH). Under iodoform conditions, ethanol is oxidized to acetaldehyde (CH3CHO), which then undergoes iodination and cleavage to give CHI3 (iodoform). Therefore, ethanol gives a positive iodoform test. Step 3: Analyze option (b) C2H5CHO (propanal). Propanal is CH3CH2CHO. This is an aldehyde but does NOT have a methyl group directly attached to the carbonyl carbon (it has an ethyl group). It does not satisfy the CH3CO- structural requirement and does not give iodoform test. Step 4: Analyze option (c) (CH2OH)2 (ethylene glycol). This is HOCH2-CH2OH. Neither carbon bearing OH has a methyl group attached, so it does not give a positive iodoform test. Step 5: Since option (a) gives a positive iodoform test, option (d) 'None of these' is incorrect. Therefore, the correct answer is A.